91Ó°ÊÓ

The mean hourly wage for manufacturing workers in the United States is $20.10 (Bureau of Labor Statistics, January 2016). It is supposed that the distribution of manufacturing wage rates nationwide can be approximated by a normal distribution with a standard deviation of $1.25 per hour.

a.

Let x be the random variable that represents the hourly wage for a nationwide job search.

X follows a normal distribution with a mean of $20.10 and a standard deviation of $1.25

$$\begin{gathered} P(x>21.40)=P\left(\frac{x-20.10}{1.25}>\frac{21.40-20.10}{1.25}\right) \\ =P(z>1.04) \\ =1-P(z≤1.04) \\ =1-0.85083 \\ =0.14917 \end{gathered}$$

So, the proportion of wage rate greater than $21.40 is 0.14917

b.

Let x be the random variable that represents the hourly wage for a nationwide job search.

X follows a normal distribution with a mean $20.10 and a standard deviation $1.25

$$\begin{gathered} P(x>21.40)=P\left(\frac{x-20.10}{1.25}>\frac{21.40-20.10}{1.25}\right) \\ =P(z>1.04) \\ =1-P(z≤1.04) \\ =1-0.85083 \\ =0.14917 \end{gathered}$$

So,, the probability of firm would pay more than $21.40 is 0.14917

c.

Let $\mathrm{η}$be the median of the random variable x.

The median is the value of the variable which divides the distribution into two equal parts.

$$\begin{gathered} P(x≤\mathrm{η})=0.50 \\ P\left(\frac{x-20.10}{1.25}≤\frac{\mathrm{η}-20.10}{1.25}\right)=0.50 \\ P\left(z≤\frac{\mathrm{η}-20.10}{1.25}\right)=0.50 \\ \mathrm{Φ}\left(\frac{\mathrm{η}-20.10}{1.25}\right)=0.50 \\ \frac{\mathrm{η}-20.10}{1.25}={\mathrm{Φ}}^{-1}(0.50) \\ \frac{\mathrm{η}-20.10}{1.25}=0 \\ \mathrm{η}-2010=0 \\ \mathrm{η}=20.10 \end{gathered}$$

So, the median value of wage is $20.10

Here the mean and median value is the same. Since the distribution of wages is normal.